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Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. (For clarification, the L-length subarray could occur before or after the M-length subarray.)
Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:
0 <= i < i + L - 1 < j < j + M - 1 < A.length, or0 <= j < j + M - 1 < i < i + L - 1 < A.length.
Example 1:
Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.
Example 2:
Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.
Example 3:
Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.
Note:
L >= 1M >= 1L + M <= A.length <= 10000 <= A[i] <= 1000
给出非负整数数组 A ，返回两个非重叠（连续）子数组中元素的最大和，子数组的长度分别为 L 和 M。（这里需要澄清的是，长为 L 的子数组可以出现在长为 M 的子数组之前或之后。）
从形式上看，返回最大的 V，而 V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) 并满足下列条件之一：
0 <= i < i + L - 1 < j < j + M - 1 < A.length, 或0 <= j < j + M - 1 < i < i + L - 1 < A.length.
示例 1：
输入：A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2 输出：20 解释：子数组的一种选择中，[9] 长度为 1，[6,5] 长度为 2。
示例 2：
输入：A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2 输出：29 解释：子数组的一种选择中，[3,8,1] 长度为 3，[8,9] 长度为 2。
示例 3：
输入：A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3 输出：31 解释：子数组的一种选择中，[5,6,0,9] 长度为 4，[0,3,8] 长度为 3。
提示：
L >= 1M >= 1L + M <= A.length <= 10000 <= A[i] <= 1000
Runtime: 24 ms
Memory Usage: 19.1 MB
1 class Solution { 2 let N:Int = 1010 3 var prefix:[Int] = [Int](repeating:0,count:1010) 4 func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int { 5 var n:Int = A.count 6 for i in 1...n 7 { 8 prefix[i] = prefix[i - 1] + A[i - 1] 9 } 10 var ans:Int = 0 11 var best_l:Int = 0 12 for i in (L + M)...n 13 { 14 best_l = max(best_l, prefix[i - M] - prefix[i - M - L]) 15 ans = max(ans, best_l + prefix[i] - prefix[i - M]) 16 } 17 var best_m:Int = 0 18 for i in (L + M)...n 19 { 20 best_m = max(best_m, prefix[i - L] - prefix[i - M - L]) 21 ans = max(ans, best_m + prefix[i] - prefix[i - L]) 22 } 23 return ans 24 } 25 }
24ms
1 class Solution { 2 func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int { 3 var leftMax = [Int](repeating: 0, count: A.count) 4 var curMax = 0, curSum = 0 5 for i in A.indices { 6 curSum += i < M ? A[i] : A[i] - A[i-M] 7 curMax = max(curMax, curSum) 8 leftMax[i] = curMax 9 } 10 11 var rightMax = [Int](repeating: 0, count: A.count+1) 12 var i = A.count - 1, ans = 0 13 curMax = 0; curSum = 0 14 while i >= 0 { 15 curSum += (i + M > A.count - 1) ? A[i] : A[i] - A[i+M] 16 curMax = max(curMax, curSum) 17 rightMax[i] = curMax 18 i -= 1 19 } 20 21 curSum = 0 22 for i in A.indices { 23 curSum += i < L ? A[i] : A[i] - A[i-L] 24 let leftMax = i >= L ? leftMax[i-L] : 0 25 ans = max(ans, max(leftMax, rightMax[i+1]) + curSum) 26 } 27 return ans 28 } 29 }
84ms
1 class Solution { 2 func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int { 3 var sumLs = [Int](repeating: 0, count: A.count) 4 var sumMs = [Int](repeating: 0, count: A.count) 5 var forwardMaxLs = [Int](repeating: 0, count: A.count) 6 var backwardMaxLs = [Int](repeating: 0, count: A.count) 7 var forwardMaxMs = [Int](repeating: 0, count: A.count) 8 var backwardMaxMs = [Int](repeating: 0, count: A.count) 9 10 var sumL = 0 11 var sumM = 0 12 for i in 0..<A.count { 13 sumL += A[i] 14 if i >= L - 1 { 15 if i >= L { 16 sumL -= A[i - L] 17 } 18 sumLs[i] = sumL 19 } 20 sumM += A[i] 21 if i >= M - 1 { 22 if i >= M { 23 sumM -= A[i - M] 24 } 25 sumMs[i] = sumM 26 } 27 } 28 var maxSumL = Int.min 29 var maxSumM = Int.min 30 for i in 0..<A.count { 31 maxSumL = max(maxSumL, sumLs[i]) 32 forwardMaxLs[i] = maxSumL 33 maxSumM = max(maxSumM, sumMs[i]) 34 forwardMaxMs[i] = maxSumM 35 } 36 maxSumL = Int.min 37 maxSumM = Int.min 38 for i in (0..<A.count).reversed() { 39 maxSumL = max(maxSumL, sumLs[i]) 40 backwardMaxLs[i] = maxSumL 41 maxSumM = max(maxSumM, sumMs[i]) 42 backwardMaxMs[i] = maxSumM 43 } 44 var maxSum = Int.min 45 for l in 0..<A.count { 46 var m = A.count - 1 47 while l <= m - M { 48 maxSum = max(maxSum, forwardMaxLs[l] + backwardMaxMs[m]) 49 m -= 1 50 } 51 } 52 for m in 0..<A.count { 53 var l = A.count - 1 54 while m <= l - L { 55 maxSum = max(maxSum, backwardMaxLs[l] + forwardMaxMs[m]) 56 l -= 1 57 } 58 } 59 return maxSum 60 } 61 }
108ms
1 class Solution { 2 func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int { 3 var prefixSums: [Int] = [0] 4 var sum = 0 5 for num in A { 6 sum += num 7 prefixSums.append(sum) 8 } 9 var res = 0 10 for lRight in L..<prefixSums.count { 11 let lSum = prefixSums[lRight] - prefixSums[lRight - L] 12 for mRight in stride(from: lRight + M, to: prefixSums.count, by: 1) { 13 let mSum = prefixSums[mRight] - prefixSums[mRight - M] 14 let total = mSum + lSum 15 res = max(res, total) 16 } 17 18 for mRight in stride(from: M, to: lRight - L, by: 1) { 19 let mSum = prefixSums[mRight] - prefixSums[mRight - M] 20 let total = mSum + lSum 21 res = max(res, total) 22 } 23 } 24 return res 25 } 26 }